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Difference between revisions of "2007 AMC 8 Problems/Problem 20"
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==Problem==
 
==Problem==
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Before district play, the Unicorns had won <math>45</math> of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Before the district play, the Unicorns had won <math>45</math>% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60 </math>
 
<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60 </math>
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==Solution 1==
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At the beginning of the problem, the Unicorns had played <math>y</math> games and they had won <math>x</math> of these games. From the information given in the problem, we can say that <math>\frac{x}{y}=0.45.</math> Next, the Unicorns win 6 more games and lose 2 more, for a total of <math>6+2=8</math> games played during district play. We are told that they end the season having won half of their games, or <math>0.5 </math> of their games. We can write another equation: <math>\frac{x+6}{y+8}=0.5.</math> This gives us a system of equations:
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<math>\frac{x}{y}=0.45</math> and <math>\frac{x+6}{y+8}=0.5.</math>
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We first multiply both sides of the first equation by <math>y</math> to get <math>x=0.45y.</math> Then, we multiply both sides of the second equation by <math>(y+8)</math> to get <math>x+6=0.5(y+8).</math> Applying the Distributive Property gives yields <math>x+6=0.5y+4.</math> Now we substitute <math>0.45y</math> for <math>x</math> to get <math>0.45y+6=0.5y+4.</math> Solving gives us <math>y=40.</math> Since the problem asks for the total number of games, we add on the last 8 games to get the solution <math>\boxed{\textbf{(A)}\ 48}</math>.
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==Solution 2 (Answer Choices)==
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We can check each answer choice from left to right to see which one is correct. Suppose the Unicorns played <math>48</math> games in total. Then, after district play, they would have won <math>24</math> games. Now, consider the situation before district play. The Unicorns would have won <math>18</math> games out of <math>40</math>. Converting to a percentage, <math>18/40 = 45</math>%. Thus, the answer is <math>\boxed{\textbf{(A)} 48}</math>.
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Note: If A didn't work, we would have similarly tested the other choices until we found one that did.
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~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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==Solution 3 (Quick)==
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We know that <math>45\%=\frac9{20}</math>. Therefore, the number of games before district play must be a multiple of <math>20</math> in order for the number of games won to be an integer. The Unicorns played <math>6+2=8</math> more games during district play. The only answer choice that is <math>8</math> more than a multiple of <math>20</math> is <math>\boxed{\textbf{(A)} 48}</math>.
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==Video Solution by OmegaLearn==
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https://youtu.be/rQUwNC0gqdg?t=1993
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~pi_is_3.14
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==Video Solution by WhyMath==
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https://youtu.be/1CAxNXM8TWo
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=19|num-a=21}}
 
{{AMC8 box|year=2007|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 19:20, 18 November 2024

Problem

Before the district play, the Unicorns had won $45$% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 50\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 60$

Solution 1

At the beginning of the problem, the Unicorns had played $y$ games and they had won $x$ of these games. From the information given in the problem, we can say that $\frac{x}{y}=0.45.$ Next, the Unicorns win 6 more games and lose 2 more, for a total of $6+2=8$ games played during district play. We are told that they end the season having won half of their games, or $0.5$ of their games. We can write another equation: $\frac{x+6}{y+8}=0.5.$ This gives us a system of equations: $\frac{x}{y}=0.45$ and $\frac{x+6}{y+8}=0.5.$ We first multiply both sides of the first equation by $y$ to get $x=0.45y.$ Then, we multiply both sides of the second equation by $(y+8)$ to get $x+6=0.5(y+8).$ Applying the Distributive Property gives yields $x+6=0.5y+4.$ Now we substitute $0.45y$ for $x$ to get $0.45y+6=0.5y+4.$ Solving gives us $y=40.$ Since the problem asks for the total number of games, we add on the last 8 games to get the solution $\boxed{\textbf{(A)}\ 48}$.

Solution 2 (Answer Choices)

We can check each answer choice from left to right to see which one is correct. Suppose the Unicorns played $48$ games in total. Then, after district play, they would have won $24$ games. Now, consider the situation before district play. The Unicorns would have won $18$ games out of $40$. Converting to a percentage, $18/40 = 45$%. Thus, the answer is $\boxed{\textbf{(A)} 48}$.

Note: If A didn't work, we would have similarly tested the other choices until we found one that did.

~cxsmi

Solution 3 (Quick)

We know that $45\%=\frac9{20}$. Therefore, the number of games before district play must be a multiple of $20$ in order for the number of games won to be an integer. The Unicorns played $6+2=8$ more games during district play. The only answer choice that is $8$ more than a multiple of $20$ is $\boxed{\textbf{(A)} 48}$.

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=1993

~pi_is_3.14

Video Solution by WhyMath

https://youtu.be/1CAxNXM8TWo

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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